You actually already said the answer. Ker(π) is isomorphic to Zn and H∩Ker(π) is a subgroup of Ker(π), and so we can consider it as a subgrouop of Zn.
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manpreet![Tuteehub forum best answer]()
Best Answer
3 years ago
In the syllabus for my course on group theory, there is the following theorem which is proven: "If H⊂ZnH⊂Zn is a subgroup, then H≅ZkH≅Zk for 0≤k≤n0≤k≤n".
I understand most of the proof, but there's a simple line that is intuitive, yet doesn't mathematically make sense to me.
In the proof, which is done by induction, the following homomorphism is defined
The kernel of the map is considered, which are all vectors of the form (m1,...,mn,0)(m1,...,mn,0).
Then the syllabus reads "Since H⊂Zn+1H⊂Zn+1 is a subgroup, also H∩ Ker
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